An object is moving in a straight line according to s(t)= -t^2+10t+30 where s(t) is how far the object goes in miles in t time in hours.
a. What is the constant acceleration of the object?
b. What is the farthest distance the object travelled from the starting point?
idk r u talking in English
s(t)= -t^2+10t+30
v(t) = -2t+10
a(t) = -2
Now you can answer the questions.
So the constant acceleration is a(t)= -2?
correct
ok. so for my question b. would I plug in the acceleration into some equation to figure out farthest distance objecy traveled?
no. You want the maximum of s(t) which occurs when v(t) = 0
v = -2t+10
v=0 when t=5
s(5) = 55
Don't forget your Algebra I now that you are taking calculus. You just want to find the vertex of a parabola. You know that for y = ax^2+bx+c that occurs when x = -b/2a
Coincidentally, since dy'dx = 2ax+b, that is where y'=0
Ok so the answer is 55 miles correct? Where are you getting the 5 from?
do you even read what I write?
v = -2t+10
v=0 when t=5
To find the constant acceleration of the object, we need to differentiate the given equation s(t) with respect to time (t) twice.
a. Constant acceleration can be found using the second derivative of s(t).
First, let's find the first derivative of s(t):
s'(t) = -2t + 10
Now, let's find the second derivative of s(t) by differentiating s'(t):
s''(t) = d/dt (-2t + 10)
= -2
Therefore, the constant acceleration of the object is -2 miles per hour squared.
b. To find the farthest distance the object traveled from the starting point, we need to find the maximum value of s(t).
The maximum (or minimum) value of a quadratic function occurs at its vertex. The vertex of a quadratic function in the form s(t) = at^2 + bt + c is given by the formula:
t = -b / (2a)
In this case, a = -1, b = 10, and c = 30. Substituting these values into the formula, we can find the time (t) at which the object reaches the farthest distance from the starting point:
t = -10 / (2 * -1)
= -10 / -2
= 5
Now, substitute the value of t back into the original equation to find the farthest distance traveled:
s(5) = -(5)^2 + 10(5) + 30
= -25 + 50 + 30
= 55
Therefore, the farthest distance the object traveled from the starting point is 55 miles.